Alexander's parents are saving for his college fund. They Which formula would be used for this situation

Which measurement is the best estimate for the mass of a penny?

Maksim231197 [3]

B is correct - 2.5g :)

8 0

4 years ago

Rhett is solving the quadratic equation 0= x2 – 2x – 3 using the quadratic formula. Which shows the correct substitution of the

lesya692 [45]

Answer:

The correct option is 1.

Step-by-step explanation:

If a quadratic equation is defined as

$ax^2+bx+c=0$

then the quadratic formula is

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

The given equation is

$0=x^2-2x-3$

Here, a=1, b=-2 and c=-3.

Substitute these values in the above formula.

$x=\frac{-(-2)\pm \sqrt{(-2)^2-4(1)(-3)}}{2(1)}$

$x=\frac{2\pm \sqrt{(-2)^2-4(1)(-3)}}{2(1)}$

Therefore, the correct option is 1.

4 0

4 years ago

FIND THE SLOPE OF THE LINE

schepotkina [342]

Answer:

4/1

Step-by-step explanation:

To get from one dot to the next, you go up by four and to the right by one, making the slope 4/1

3 0

3 years ago

Last year it was found that on average it took students 20 minutes to fill out the forms required for graduation. This year the

Darina [25.2K]

Answer:

$t=\frac{18.5-20}{\frac{5.2}{\sqrt{22}}}=-1.35$

$p_v =P(t_{21}$

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the time is not significantly less than 20 minutes at 5% of significance.

Step-by-step explanation:

Data given and notation

$\bar X=18.5$ represent the sample mean

$s=5.2$ represent the standard deviation for the sample

$n=22$ sample size

$\mu_o =20$ represent the value that we want to test

$\alpha$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the mean for the new forms take less time to complete than the older forms, the system of hypothesis would be:

Null hypothesis: $\mu \geq 20$

Alternative hypothesis: $\mu < 20$

Compute the test statistic

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

We can replace in formula (1) the info given like this:

$t=\frac{18.5-20}{\frac{5.2}{\sqrt{22}}}=-1.35$

Now we need to find the degrees of freedom for the t distirbution given by:

$df=n-1=22-1=21$

What do you conclude?

Compute the p-value

Since is a one left tailed test the p value would be:

$p_v =P(t_{21}$

If we compare the p value and a significance level assumed $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the actual true mean for the time is not significantly less than 20 minutes at 5% of significance.

7 0

3 years ago

Can you please answers my question

aev [14]

The Answer is 5 groups

6 0

3 years ago